package com.xypower.link;

import java.util.Stack;

public class IsPalindromeList {
    public static class Node{
        public int value;
        public Node next;
        public Node(int data){
            this.value = data;
        }
    }
    //判断一个链表是否是回文结构 1->2->3->2->1->null
    // N额外空间 解法1
    public static boolean isPalindrome1(Node head){
        Stack<Node> stack = new Stack<Node>();
        Node cur =  head;
        while (cur!=null){
            stack.push(cur);
            cur = cur.next;
        }
        while (head!=null){
            if (head.value!=stack.pop().value){
                return false;
            }
            head = head.next;
        }
        return true;
    }
    // N/2额外空间 解法2
    public static boolean isPalindrome2(Node head){
        if (head ==null ||head.next == null){
            return true;
        }
        //快慢指针
        Node right = head.next;
        Node cur = head;
        while(cur.next!=null&&cur.next.next!=null){
            right = right.next;
            cur = cur.next.next;
        }
        Stack<Node> stack = new Stack<Node>();
        while(right!=null){
            stack.push(right);
            right =right.next;
        }
        while (!stack.isEmpty()){
            if (head.value!=stack.pop().value){
                return false;
            }
            head = head.next;
        }
        return true;
    }

    // o(1)额外空间 解法3
    public static boolean isPalindrome3(Node head){
        if (head ==null ||head.next == null){
            return true;
        }
        //快慢指针
        Node n1 = head;
        Node n2 = head;
        while (n2.next!=null && n2.next.next!=null){
            n1 = n1.next;
            n2 = n2.next.next;
        }
        n2 = n1.next;
        n1.next = null;
        Node n3 = null;
        while (n2 != null){
            n3 = n2.next;
            n2.next = n1;
            n1 = n2;
            n2 = n3;
        }
        //开始遍历两边往中间遍历
        n3 = n1;
        n2 = head;
        boolean res = true;
        while (n1 != null && n2 != null){
            if (n1.value!=n2.value){
                res = false;
                break;
            }
            n1 = n1.next;
            n2 = n2.next;
        }
        //右部分逆序回来
        n1 = n3.next;
        n3.next = null;
        while (n1 != null){
            n2 = n1.next;
            n1.next = n3;
            n3 = n1;
            n1 = n2;
        }
        return res;
    }
}
